How to show that $(\lambda_n x_n)$ is compact iff $(\lambda_n)\in c_0$?

Question

Let $(\lambda_n) \in l_\infty$, and $A:l_2 \rightarrow l_2$ a linear operator defined as $$A(x_n)=(\lambda_n x_n), \,\, (x_n) \in l_2$$ How to show that $A$ is compact if and only if ($\lambda_n)\in c_0$?

Answer 1

One way: if $\lambda_n \to 0$, we can approximate $A$ in norm by finite-rank operators (replacing $\lambda_n$ for $n > N$ by $0$).

Other way: if $\lambda_n$ does not go to $0$, find a sequence (a subsequence of the standard unit vectors) whose image under $A$ has no convergent subsequence.